In (13x+2)/(x^3-1)13x+2x3−1, we should first factorize x^3-1x3−1, using identity (a^3+b^3)=(a+b)(a^2-ab+b^2)(a3+b3)=(a+b)(a2−ab+b2).
Hence factors are given by (x^3-1)=(x-1)(x^2-x+1)(x3−1)=(x−1)(x2−x+1).
Partial fractions of (13x+2)/(x^3-1)13x+2x3−1, will be
(13x+2)/(x^3-1)hArrA/(x-1)+(Bx+C)/(x^2+x+1)13x+2x3−1⇔Ax−1+Bx+Cx2+x+1and simplifying RHS it becomes
(13x+2)/(x^3-1)hArr(A(x^2+x+1)+(Bx+C)(x-1))/((x-1)(x^2+x+1))13x+2x3−1⇔A(x2+x+1)+(Bx+C)(x−1)(x−1)(x2+x+1) or
(13x+2)/(x^3-1)hArr((Ax^2+Ax+A)+(Bx^2-Bx+Cx-C))/(x^3-1)13x+2x3−1⇔(Ax2+Ax+A)+(Bx2−Bx+Cx−C)x3−1 or
(13x+2)/(x^3-1)hArr((A+B)x^2+(A-B+C)x+(A-C))/(x^3-1)13x+2x3−1⇔(A+B)x2+(A−B+C)x+(A−C)x3−1
Hence, we have A+B=0A+B=0 - (i), A-B+C=13A−B+C=13 - (ii) and A-C=2A−C=2 - (iii).
From (i) we get B=-AB=−A, from (iii) C=A-2C=A−2 and putting them in (ii),
we get A-(-A)+(A-2)=13A−(−A)+(A−2)=13 or 3A=153A=15 or A=5A=5
Hence B=-5B=−5 and C=3C=3
Hence partial fractions are 5/((x-1))+(-5x+3)/(x^2+x+1)5(x−1)+−5x+3x2+x+1
